13. x11+2x1 +26x = 600 cos 10t; x(0) = 10, x1(0) = 0
To solve these problems, we need to find both the steady periodic solution xsp(t)x_{\text{sp}}(t)xsp(t) and the actual solution x(t)x(t)x(t), which is a combination of the steady periodic solution and a transient term xtr(t)x_{\text{tr}}(t)xtr(t). Here's the general approach for each problem:
1. Steady Periodic Solution xsp(t)x_{\text{sp}}(t)xsp(t)
The given differential equations are likely non-homogeneous, and the steady periodic solution corresponds to the particular solution that oscillates at the driving frequency. It will be of the form:
xsp(t)=Ccos(ωt−α)x_{\text{sp}}(t) = C \cos(\omega t - \alpha)xsp(t)=Ccos(ωt−α)
To find CCC and α\alphaα:
- Convert the forcing function into a form with sinusoidal terms.
- Use the method of undetermined coefficients or phasor analysis to match the amplitude and phase with the forcing function.
- This will give you the steady-state response to the external forcing term.
2. Transient Solution xtr(t)x_{\text{tr}}(t)xtr(t)
The transient solution xtr(t)x_{\text{tr}}(t)xtr(t) solves the homogeneous equation. Typically, the general solution to the homogeneous part is:
xtr(t)=Aeλ1t+Beλ2tx_{\text{tr}}(t) = A e^{\lambda_1 t} + B e^{\lambda_2 t}xtr(t)=Aeλ1t+Beλ2t
where λ1\lambda_1λ1 and λ2\lambda_2λ2 are the characteristic roots. For second-order linear equations, these roots depend on the damping and can represent:
- Overdamping: real and distinct roots,
- Critical damping: real and repeated roots,
- Underdamping: complex conjugate roots.
3. Actual Solution x(t)x(t)x(t)
The actual solution is the sum of the steady-state and transient solutions:
x(t)=xsp(t)+xtr(t)x(t) = x_{\text{sp}}(t) + x_{\text{tr}}(t)x(t)=xsp(t)+xtr(t)
4. Applying Initial Conditions
To determine the constants AAA and BBB, apply the given initial conditions x(0)x(0)x(0) and x′(0)x'(0)x′(0) to the actual solution x(t)x(t)x(t) and solve for the unknowns.
13. x"+2x'+26x=600cos10t; x(0)=10, x'(0)=0
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